Calculus
posted by Anonymous .
f(x)=(5x)/(1+2x^2).
f'(4)=75/289
f'(x)=(55x^2)/1+2x^2+x^4)
Use this equation of the tangent line to the curve y=5x/(1+x^2) at the point (4, 1.17647). The equation of the tangent line can be written in the form y=mx+b. m=75/289. What is b?
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